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Published on: 24/08/2019
Inverse Trigonometric Functions
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Take MCQ Maths Test
1.
Show that \({ sin }^{ -1 }\left( 2x\sqrt { 1-{ x }^{ 2 } } \right) =2{ sin }^{ -1 }x\) ,\(-\frac{1}{\sqrt{2}} \leq x \leq \frac{1}{\sqrt{2}}\)
2.
Evaluate : \( \sin { \left( 2\cos ^{ -1 }{ \left( -\frac { 3 }{ 5 } \right) } \right) } \)
3.
Write the value of cot (tan-1a + cot-1a).
4.
What is principal value of \(tan^{ -1 }\left( tan\frac { 2\pi }{ 3 } \right) \) ?
5.
Write the principal value of \({ tan }^{ -1 }\left( tan\frac { 7\pi }{ 6 } \right) \)
6.
Find the values of each of the expressions in Exercises : \(\tan ^{-1}\left(\tan \frac{3 \pi}{4}\right)\)
7.
Find the principal values of the following: \({ sec }^{ -1 }\left( \frac { 2 }{ \sqrt { 3 } } \right) \)
8.
if (a < 0) and x \(\varepsilon \) (-a, a), simplify tan-1 \(\left( \frac { x }{ \sqrt { { a }^{ 2 }-{ x }^{ 2 } } } \right) \)
9.
if tan-1 (a/x) + tan-1 (b/x) = \(\pi\) /2, then x =
1.
Let x = sin θ. Then sin–1 x = θ. We have
\( \sin ^{-1}\left(2 x \sqrt{1-x^2}\right) =\sin ^{-1}\left(2 \sin \theta \sqrt{1-\sin ^2 \theta}\right) \)
\( =\sin ^{-1}(2 \sin \theta \cos \theta)=\sin ^{-1}(\sin 2 \theta)=2 \theta \)
\( =2 \sin ^{-1} x\)
2.
\( \sin { \left( 2\cos ^{ -1 }{ \left( -\frac { 3 }{ 5 } \right) } \right) } \)
\(=2\sin { \left( 2\cos ^{ -1 }{ \left( -\frac { 3 }{ 5 } \right) } \right) } cos\left( { cos }^{ -1 }\left( -\frac { 3 }{ 5 } \right) \right) \)
\(\\ =2\sin { \left( \cos ^{ -1 }{ \left( -\frac { 3 }{ 5 } \right) } \right) } \left( -\frac { 3 }{ 5 } \right) \)
\(=-2\sin { \left( \sin ^{ -1 }{ \left( \frac { 4 }{ 5 } \right) } \right) } \left( -\frac { 3 }{ 5 } \right) \)
\(=-2\left( \frac { 4 }{ 5 } \right) \left( -\frac { 3 }{ 5 } \right) \)
\(=+\frac { 24 }{ 25 } \)
3.
\(\cot \left(\tan ^{-1} a+\cos ^{-1}\right) \)
\(=\cot \left(\frac{\pi}{2}\right) \quad\left[\tan ^{-1} x+\cot ^{-1} x=\frac{\pi}{2}\right] \)
\(=0\)
4.
\(tan^{ -1 }\left( tan\frac { 2\pi }{ 3 } \right) ={ tan }^{ -1 }\left( -tan\frac { \pi }{ 3 } \right) \)
\(=-\frac { \pi }{ 3 } \)
5.
\({ tan }^{ -1 }\left( tan\frac { 7\pi }{ 6 } \right) =\frac { \pi }{ 6 } \)
Alternative Method :
\({ tan }^{ -1 }\left( tan\frac { 7\pi }{ 6 } \right) ={ tan }^{ -1 }\left[ tan\left( \pi +\frac { \pi }{ 6 } \right) \right] \)
\(={ tan }^{ -1 }\left[ tan\frac { \pi }{ 6 } \right] =\frac { \pi }{ 6 } \)
\(\left[ \because { tan }^{ -1 }(tan\theta )=\theta \forall \theta \in \left( -\frac { \pi }{ 2 } ,\frac { \pi }{ 2 } \right) \right] \)
6.
Given \(\tan ^{-1}\left(\tan \frac{3 \pi}{4}\right)\)
\(\frac{3 \pi}{4} \text { as } \frac{(4 \pi-\pi)}{4} \text { or } \pi-\frac{\pi}{4}\)
After substuting we have,
\(\tan ^{-1}\left(\tan \left(\frac{3 \pi}{4}\right)\right)=\tan ^{-1}\left(\tan \left(\pi-\frac{\pi}{4}\right)\right)=-\frac{\pi}{4}\)
7.
Let \( y=\sec ^{-1}\left(\frac{2}{\sqrt{3}}\right) \)
\(\Rightarrow \sec y=\frac{2}{\sqrt{3}} \)
\(\Rightarrow \sec y=\sec \left(\frac{\pi}{6}\right)\)
We know that the range of the principal value branch of \(\sec ^{-1} x \text { is }[0, \pi]-\left\{\frac{\pi}{2}\right\}\)
Hence, the principal value of \(\sec ^{-1}\left(\frac{2}{\sqrt{3}}\right) \text { is } \frac{\pi}{6}\)
8.
( )
\(-sin^{ -1 }\left( \frac { x }{ a } \right) \)
9.
( )
\(\sqrt { ab } \)
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